X forms an oxide X2O3 , 0.36 grams of X forms 0.56 grams of X2O3 .so that atomic weight of X is
let M be moles of X2O3
so in M moles of X2O3, you have
2 M of X2, and 3 M of O
so if mass X in 2M is .36, and mass O in 3M is .20, then
M=(.2/3)/18=.2/54=1/270 moles
but also
M=.18/atomicwtX
or atomicwtX=.18*270=48.6
To find the atomic weight of element X, we can use the concept of the law of definite proportions.
The law of definite proportions states that a given chemical compound always contains the same proportion of elements by mass.
Since we know that 0.36 grams of X forms 0.56 grams of X2O3, we can set up a proportion:
0.36 g X / X atomic weight = 0.56 g X2O3 / (2 × X atomic weight + 3 × (16 atomic weight))
Simplifying the equation:
0.36 g X = 0.56 g X2O3 / (2X + 48)
Cross multiplying:
0.36 g X × (2X + 48) = 0.56 g X2O3
0.72X + 17.28 = 0.56 g X2O3
Subtracting 17.28 from both sides:
0.72X = 0.56 g X2O3 - 17.28
0.72X = -16.72
Dividing both sides by 0.72:
X = -16.72 / 0.72
X ≈ -23.22
It appears that there might be an error in the values given in the question, as the calculated atomic weight of X is negative, which is not possible. Please double-check the given values or information.
To find the atomic weight of element X, we need to calculate the molar mass of X2O3 (the oxide formed by element X) using the given information.
The molar mass of X2O3 can be calculated by adding the atomic masses of 2 atoms of X and 3 atoms of oxygen (O).
Let's denote the atomic weight of X as "M".
The molar mass of X2O3 would be:
(M x 2) + (16 x 3) = M + 48 grams/mol
Now, let's calculate the molar mass of X2O3 using the given data:
0.36 grams of X forms 0.56 grams of X2O3
Therefore, the molar mass of X2O3 is:
0.56 grams / 0.36 grams/mol = 1.56 grams/mol
From our earlier equation, we know that the molar mass of X2O3 is M + 48 grams/mol.
Equating the two expressions, we get:
M + 48 grams/mol = 1.56 grams/mol
Now, subtracting 48 grams/mol from both sides:
M = (1.56 grams/mol) - 48 grams/mol
M = -46.44 grams/mol
However, since atomic weights cannot be negative, it is likely there was a calculation error or an error in the given data. Please double-check the information provided to find the correct atomic weight of element X.
Calculate % each element.
%X = (0.36/0.56)*100 = 64.3%
%O = (0.2/0.56)*100 = 35.7%
Take 100 g sample to give
64.3 g X
35.7 g O
Convert to mols.
64.3/at mass X = ?mols X.
35.7/16 = 2.23 mols O.
We know this is in the ratio of 2:3 so, letting X = atomic mass X
64.3/X = ?......*some # || = 2
35.7/16 = 2.23..*some # || = 3
What is some #? It is
2.23*# = 3. That # must be 3/2.23 = 1.34. The number multiplier for metal X must also be 1.34 so ? must be 2/1.34 = 1.49 so 64.3/X = 1.49 and X (atomic mass X) must be 64.3/1.49 = 43.2.
CHECK:
0.36g X/43.2 = 0.008333 mols X
0.2g O/16 = 0.0125 mols O
Divide both numbers by the smaller number to obtain:
0.00833/0.00833 = 1
0.0125/0.00833 = 1.50 and these are in the ratio of small whole numbers of 2:3 so formula is X2O3. The 43.2 number checks out for atomic mass X OK.
Bob Pursley obtained slightly higher in his work above which probably is the result of a typo or a rounding error. You can check it another way and see if the percentages agree.
X2O3 molar mass is (2*43.2)+3(16) = 134.4
%X = [2*(43.2)/134.4]*100 = 64.28 which rounds to 64.3 and
%O = [3*16/134.4]*100 = 35.71 which rounds to 35.7. Formula checks and percent checks; must be right.